EC Cryptography Tutorials - Herong's Tutorial Examples - Version 1.00, by Dr. Herong Yang
Reduced Elliptic Curve Group - E127(-1,3)
This section provides an example of a reduced Elliptic Curve group E127(-1,3). An example of addition operation is also provided.
Let's take a look at another reduced elliptic curve group, E127(-1,3), as discussed in "Elliptic Curve Cryptography: finite fields and discrete logarithms" by Andrea Corbellini at andrea.corbellini.name/2015/05/23/elliptic-curve-cryptography-finite-fields-and-discrete-logarithms/.
Here is the reduced elliptic curve group using modular arithmetic of prime number 127, E127(-1,3):
y2 = x3 - x + 3 (mod 127)
The above diagram provides all points in this group. It also illustrates an example of the reduced addition operation:
Given two points on the curve: P = (16,20) Q = (41,120) Draw a straight line passing through A and B, And wrap the line around when it reaches the boundary of the region. It will reach another point -R on the curve. Take the symmetric point R of -R: R = A + B R = (86,81)
Let's verify R = P+Q using algebraic equations given by the reduced elliptic curve group definition:
For any two given points on the curve: P = (xP, yP) = (16,20) Q = (xQ, yQ) = (41,120) R = P + Q is a third point on the curve: R = (xR, yR) Where: xR = m2 - xP - xQ (mod p) (11) yR = m(xP - xR) - yP (mod p) (12) m(xP - xQ) = yP - yQ (mod p) (18) Calculation: m*(16-41) = 20-120 (mod 127) m*(-25) = -100 (mod 127) 102*m = 27 (mod 127) m = 27 * 1/102 (mod 127) m = 27*66 (mod 127) m = 1782 (mod 127) m = 4 xR = 4*4 - 16 - 41 = -41 (mod 127) xR = 86 yR = 4*(16 - 113) - 20 = -300 (mod 127) yR = 81 C = (86,81)
The result from algebraic equations matches the geometrical result!
Last update: 2019.
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